Comparing two MySQL tables to populate a dropdown - php

Question

I'm trying to populate a dropdown menu by comparing two tables, one has a list of supervisor and employee numbers, the other has employee numbers and names. I need to take the numbers for each supervisor and employee and turn them into employee names for the drop down menu, so basically
TABLE payroll_employeelist
Supervisor Employee
1234 3456
1234 2239
1234 123
2910 338
2910 3901
TABLE payroll_users
number name
3456 John Smith
2239 Mary Jane
123 Joe Brown
etc
Supervisors are identified by a session variable callede $usernumber.
What I have so far and is returning one result (just one!) is the following:
if ($loademployees == 1){
echo "<option value=\"base\">---- Employee Name ----</option>";
$query = "SELECT payroll_employeelist.employee, payroll_users.number, payroll_users.name FROM payroll_employeelist WHERE supervisor='$usernumber' LEFT JOIN payroll_users ON payroll_employeelist.employee=payroll_users.number ";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
echo "<option value=\">" . $row{'name'} . "</option>";
}
echo "</select><br>";
}
Can anyone help with this? I get the feeling I've done something funny with the JOIN. It should look like a list of employee names in the dropdown.
UPDATE:
What I have now is:
if ($loademployees == 1){
echo "<option value=\"base\">---- Employee Name ----</option>";
$query = "SELECT payroll_employeelist.supervisor, payroll_employeelist.employee, payroll_users.number, payroll_users.name
FROM payroll_employeelist
INNER JOIN payroll_users
ON payroll_employeelist.employee = payroll_users.number
WHERE supervisor = '$usernumber' ";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
echo "<option value=\">" . $row['name'] . "</option>";
}
echo "</select><br>";
}
This is successfully returning one of the three records in the test data set, just one, the middle record. The $usernumber is generated internally by the way, no injection possible.
LAST UPDATE- SOLVED
The problem believe it or not was
echo "</select><br>";
it was echoing that before echoing the results of the while loop so it thought the options list was empty. I can't explain the randomly appearing single employee mind you, but it's working now.

Answer 1

You need to join payroll_users twice on table payroll_employeelist since there are two columns that are dependent on it.
SELECT sup.Name SupervisorName,
empName EmployeeName
FROM payroll_employeelist a
INNER JOIN payroll_users sup
ON a.Supervisor = sup.number
INNER JOIN payroll_users emp
ON a.Employee = emp.Number
WHERE sup.Supervisor = '$usernumber'
As a sidenote, the query is vulnerable with SQL Injection if the value(s) of the variables came from the outside. Please take a look at the article below to learn how to prevent from it. By using PreparedStatements you can get rid of using single quotes around values.
How to prevent SQL injection in PHP?

Answer 2

That's probably because you put the join condition inside the WHERE, but you probably meant to have it outside:
SELECT payroll_employeelist.employee, payroll_users.number, payroll_users.name
INNER JOIN payroll_users ON payroll_employeelist.employee=payroll_users.number
FROM payroll_employeelist
WHERE supervisor='$usernumber'
Also, if you use LEFT JOIN you will also get employees that are not attached to any user. A few things about escaping:
$query = "SELECT ... WHERE supervisor='$usernumber' ... ";
That's susceptible to SQL injection if $usernumber comes from a web request; consider using either mysql_real_escape_string() to escape it or switch to PDO / mysqli and use prepared statements instead.
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
echo "<option value=\">" . $row{'name'} . "</option>";
}
You should escape $row['name'] as well, and you shouldn't use curly braces either:
echo "<option value=\">" .
htmlspecialchars($row['name'], ENT_QUOTES, 'UTF-8') .
"</option>";

Answer 3

SELECT
tb1.supervisor,
tb1.employee,
tb2.name
FROM
payroll_employeelist AS tb1
INNER JOIN payroll_users AS tb2
ON tb1.employee = tb2.number
As you want exact matches, without non-matching values, you need INNER JOIN instead of LEFT JOIN

Answer 4

Use
SELECT payroll_employeelist.employee, payroll_users.number, payroll_users.name FROM payroll_employeelist LEFT JOIN payroll_users ON payroll_employeelist.employee = payroll_users.number WHERE supervisor = '$usernumber'
instead.

Answer 5

You should look at using PDO for your queries. Since you are dynamically assigning values into your query PDO will be a bit more secure, and if you're running this query multiple times it will be faster with PDO.
As for your query, you have your SQL Clauses ordered incorrectly. Perhaps these links will help:
PDO Tutorial from MySQL
SQL Join Tutorial

Answer 6

you have a mistake in your sintax
please check
$query = "SELECT payroll_employeelist.employee, payroll_users.number, payroll_users.name FROM payroll_employeelist WHERE supervisor='$usernumber' LEFT JOIN payroll_users ON payroll_employeelist.employee=payroll_users.number ";
should be
$query = "SELECT payroll_employeelist.`employee`, payroll_users.`number`, payroll_users.`name` FROM `payroll_employeelist` LEFT JOIN `payroll_users` ON payroll_employeelist.employee` = payroll_users.`number` WHERE `supervisor` = '$usernumber' ";
about this
WHERE `supervisor` = '$usernumber' ";
what table does supervisor is in? you need to fix with prefix payroll_employeelist or payroll_users
documentation here
I would like to also to remember you that mysql_ functions are deprecated so i would advise you to switch to mysqli or PDO for new projects.